Submit a specific form with JQuery Ajax

问题:

I had many forms with the same id on a page. When I press the submit button of any form, first the first will submit, after the second click the second ... . But I want when I press the submit button, the form will submit where the button belongs to. How can I do that.

Here my JS Code:

$(document).on('submit','#ajax_form',function(e) {
   var form = $('#ajax_form');
   var data = form.serialize();
   $.post('game/write.php', data, function(response) {
      console.log(response);
      $('#power').replaceWith(response);
   });
   return false;
});

Here the HTMl Code:

<div id="power">
<div class="span4">
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="1" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="4" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="7" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
</div>
<div class="span4">
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="2" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="5" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="8" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
</div>
<div class="span4">
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="3" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="6" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="9" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
</div>

回答1:

This is where your problem begins:

var form = $('#ajax_form'); 

It selects the first form, not the one that was submitted. Simply replacing it with

var form = $(this);

would solve your problem, but I still suggest not using duplicate id's.

回答2:

If you're not bothered about know which form is being submitted and simply want to handle the submit for all of them with one piece of code, then this will do it...

$(document).on('submit','form',function(e) {
    var form = $(this);
    var data = form.serialize();
    $.post('game/write.php', data, function(response) {
        console.log(response);
        $('#power').replaceWith(response);
    });
    return false;
});


<div id="power">
<div class="span4">
    <form action="game/write.php" method="post"><input type="hidden" value="1" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" method="post"><input type="hidden" value="4" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" method="post"><input type="hidden" value="7" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
</div>
<div class="span4">
    <form action="game/write.php" method="post"><input type="hidden" value="2" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" method="post"><input type="hidden" value="5" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" method="post"><input type="hidden" value="8" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
</div>
<div class="span4">
    <form action="game/write.php" method="post"><input type="hidden" value="3" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" method="post"><input type="hidden" value="6" name="button"><button type="submit" class="btn btn-block btn-success"><img  src="images/null.png"></button></form>
    <form action="game/write.php" method="post"><input type="hidden" value="9" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
</div>

As everyone has previously mentioned, you can't use an ID for more than one element. The above code assigns the submit event handler to all forms, and uses $(this) to reference the form that was submitted. It should do the trick :)

回答3:

Try this:

$(document).on('click','button.btn',function(e) { 
    //you will trigger this function when you click a button
    //this will select the parent, i.e., the form
    var form = $(this).parent();

    var data = form.serialize();
    $.post('game/write.php', data, function(response) {
        console.log(response);
        $('#power').replaceWith(response);
    });
    return false;
});

回答4:

ID's must be unique across the entire DOM.

回答5:

Only one element can have the same id. Set your function up to pass the id of the correct form you want to submit.

回答6:

Better use class="ajax_form" instead of ID, then apply $(this).

$(document).on('submit','.ajax_form',function(e) {
   var form = $(this);
   var data = form.serialize();
   // other code
   return false;
});